How many grams of potassium iodide are required to make 500. milliliters of a 2.80 M KI solution.

232.4 grams of KI solution

Explanation:

Well, we first start off by using like units. The concentration (M) of KI can be rewritten as Moles/ Liter, which makes the solution rewritten as 2.80 Moles/ 1 L.

We also know that 1000 Milliliters are equal to 1 Liter, therefore we can rewrite the 500 Milliliters as 0.500 Liters from doing a conversion (500 mL x  (1 L/1000 mL)= 0.500 L).

From here we multiply the concentration by the number of liters to leave us with moles ((2.80 Moles/ 1 L)  x  0.500 L= 1.4 Moles).

Now, we must find the molar mass of the KI solution, this can be done either by googleing it or adding the molar mass of each element in the compound via the periodic table. In this case, the molar mass for the Potassium in the compound is 39 g/mol while the Iodine is 127 g/mol. Those add to 166 g/mol, and now we can convert to grams from moles.

Now, we multiply the moles by the molar mass to cancel out the moles and leave us with grams. (1.4 mol x (166 g/mol)= 232.4 grams).

I hoped this helps and good luck with your semester in Chemistry!

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